b
(b) Centripetal force $=$ breaking force
$⇒\ m{\omega ^2}r = $ breaking stress $\times$ cross sectional area
$⇒\ m{\omega ^2}r = p \times A$
$⇒\ \omega = \sqrt {\frac{{p \times A}}{{mr}}} = \sqrt {\frac{{4.8 \times {{10}^7} \times {{10}^{ - 6}}}}{{10 \times 0.3}}} $
$\omega = 4\,rad/\sec $